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Edit: Atomic orbitals For atoms you have a system that is spherically symmetric. Thus, it is most conveniently expressed in spherical coordinates instead of cartesian ones. … $\mathrm{s}$ orbitals are spheres while $\mathrm{p}$ orbitals are dumbbell-shaped (side note: the Spherical Harmonics are complex functions which are rather hard to visualize; the shapes that every chemist …

Another reason is, that the $3 \ce{p}$ or $4 \ce{p}$ orbitals are much more diffuse than the $2 \ce{p}$ orbitals. If you want to read deeper into the topic have a look at: [1] W. Kutzelnigg, Angew. …

No it does not matter whether you start with 'spin up' or 'spin down' electrons as long as you are consistent: If you start filling some orbitals with 'spin up' electrons you will have to keep doing this … for all the orbitals you want to fill. …

Firstly, you seem to have a misconception about atomic orbitals and molecular orbitals. … you use these orbitals to form bonds between carbon and, say, the 1s orbitals of four hydrogen atoms thus getting methane $\ce{CH4}$. …

When combined at a given atomic center, any two atomic orbitals add in a vectorial manner. … The mixing of atomic orbitals with different angular momentum quantum number is also controlled by a vectorial addition and leads to various types of hybrid orbitals some examples of which are shown here …

You are attaching too much importance to Lewis structures. The 8-electron rule and Lewis structures which are derived from it are only rough guidelines for working out the electronic structure of a co …

You should definitely post the MO diagram of $\ce{F2}$ that you worked out. When trying to solve such a problem you should try to analyse the question as thoroughly as possible to make sure that you c …

Have a look here: Orbitals that have the same value of the principal quantum number $n$ form a shell. … With orbitals in the context of shells and subshells one usually means atomic orbitals, i.e. two-electron eigenstates of an atom's Hamilton operator which are characterized by the three quantum numbers …

& 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & & \\ \mathrm{E_u} & 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 & (x,y) & \\ \hline \end{array}$$ Finally you need to determine the symmetry labels of the p- and d-orbitals … \end{align}$$ The five d-orbitals show the same transformational properties as their quadratic function counterparts and thus they get the symmetry labels, i.e. …

\end{align} If you evaluate this integral for the $2\ce{s}$ and $2\ce{p}$ orbitals you get \begin{align} \langle r \rangle_{2\ce{s}} = \langle r \rangle_{2, 0} &= \frac{6 a_{0}}{ Z } \\ \langle …

For hydrogen atom, L^2 and Lz can be obtained as eigenvalues for a particular wave function. But that does not completely specify the angular momentum vector. How to get about this problem? Well, …

I think your question implicates another question (which is also mentioned in some comments here), namely: Why are all energy eigenvalues of states with a different angular momentum quantum number $\e …

which in turn leads to a cascade of contractions of those outer s orbitals. … chemical reactions and they are harder to remove via ionization, because the s orbitals' decreased size lessens the orbital overlap with potential reaction partners' orbitals and leads to a lower energy …

General case There is indeed a mathematical theorem that deals with the number of nodes an eigenfunction corresponding to a certain eigenvalue can possess. It was laid down by Courant$^{[1, 2]}$ and …

They interact with the filled high-lying neighboring $\ce{C-H}$ or $\ce{C-C}$ $\ce{\sigma}$ bonds in $\pi$-bond-like manner and so from those two fragment orbitals you will get two molecular orbitals … {C=C})$ with $\sigma (\ce{C-X})$, which will be higher in energy then the fragment orbitals. …